Understanding entropy generation

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In thermodynamics, when work is done on a closed system, such as with a paddle wheel, entropy generation (Sgen) can be non-zero despite no entropy transfer occurring. In adiabatic processes, the work done can lead to changes in entropy that are entirely attributed to Sgen. Quasi-static processes are often considered reversible, and in such cases, Sgen is typically zero. It is generally safe to assume Sgen is zero in reversible processes, but caution is advised when heat transfer occurs between objects with differing thermal properties. Understanding these nuances is crucial for accurate thermodynamic analysis.
Urmi Roy
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Hi all,

I've been doing some pretty complex problems on closed systems in thermodynamics, and a few questions pooped up in my mind about Sgen (entropy generation). Please help me out with them..

1. When work is done on a system (like a paddle wheel in a cylinder containing gas), there is no entropy 'transfer' but is Sgen non zero? Then the 'entropy change' that we get from the constitutive relations for entropy change is wholly Sgen?

An interesting example of this an adiabatic process, where an insulated piston-cylinder system experiences work from outside, and the gas is compressed.

2. Is quasi-static equivalent to reversible? Is Sgen zero for quasi-static/reversible process?

3. Under what circumstances is it safe to assume entropy generation is 0? I usually assume so when there is heat transfer between 2 objects that have very different thermal conductivity or hear capacities...is that ok? Is there any additional situation?

Thanks!
-Urmi
 
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i think Sgen is zero in a completely reversible process. S2-S1= dQ/T +Sgen
 

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